States of Matter Practice Questions - DPP for JEE

 Page 1


1. (a) From the graph we can see the correct order of pressures  p
1
 > p
3 
>
p
2
2. (b) Gases become cooler during Joule Thomson’s expansion only if they are
below a certain temperature known as inversion temperature (T
i
). The
inversion temperature is characteristic of each gas and is given by
,   where R is gas constant
Given a = 0.244 atm L
2
 mol
–2
b = 0.027 L mol
–1
R = 0.0821 L atm deg
–1
 mol
–1
? 
3. (d)
4. (d) Let the mass of methane and oxygen = m gm.
Mole fraction of O
2
= 
=  =  = 
Partial pressure of O
2
 = Total pressure × mole fraction of O
2
, = P × 
 = 
5. (c) The expression of root mean square speed is
Hence,
Page 2


1. (a) From the graph we can see the correct order of pressures  p
1
 > p
3 
>
p
2
2. (b) Gases become cooler during Joule Thomson’s expansion only if they are
below a certain temperature known as inversion temperature (T
i
). The
inversion temperature is characteristic of each gas and is given by
,   where R is gas constant
Given a = 0.244 atm L
2
 mol
–2
b = 0.027 L mol
–1
R = 0.0821 L atm deg
–1
 mol
–1
? 
3. (d)
4. (d) Let the mass of methane and oxygen = m gm.
Mole fraction of O
2
= 
=  =  = 
Partial pressure of O
2
 = Total pressure × mole fraction of O
2
, = P × 
 = 
5. (c) The expression of root mean square speed is
Hence,
6. (a) Given T = 27°C = 27 + 273 = 300 K
V = 10.0 L
Mass of He = 0.4 g
Mass of oxygen = 1.6 g
Mass of nitrogen = 1.4 g
n He = 0.4/4 = 0.1
n O
2
 = 1.6/32 = 0.05
n N
2
 = 1.4/28 = 0.05
n total = n He + n O
2
 + n N
2
 = 0.1 + 0.05 + 0.05 = 0.2
P =  atm
7. (b) Rate of diffusion ?
 Molecular mass of HCl > Molecular mass of NH
3
 HCl diffuses at slower rate and white ammonium chloride is
first formed near HCl bottle.
8. (b)    
Mean molar mass = 
= 
9. (a)
Page 3


1. (a) From the graph we can see the correct order of pressures  p
1
 > p
3 
>
p
2
2. (b) Gases become cooler during Joule Thomson’s expansion only if they are
below a certain temperature known as inversion temperature (T
i
). The
inversion temperature is characteristic of each gas and is given by
,   where R is gas constant
Given a = 0.244 atm L
2
 mol
–2
b = 0.027 L mol
–1
R = 0.0821 L atm deg
–1
 mol
–1
? 
3. (d)
4. (d) Let the mass of methane and oxygen = m gm.
Mole fraction of O
2
= 
=  =  = 
Partial pressure of O
2
 = Total pressure × mole fraction of O
2
, = P × 
 = 
5. (c) The expression of root mean square speed is
Hence,
6. (a) Given T = 27°C = 27 + 273 = 300 K
V = 10.0 L
Mass of He = 0.4 g
Mass of oxygen = 1.6 g
Mass of nitrogen = 1.4 g
n He = 0.4/4 = 0.1
n O
2
 = 1.6/32 = 0.05
n N
2
 = 1.4/28 = 0.05
n total = n He + n O
2
 + n N
2
 = 0.1 + 0.05 + 0.05 = 0.2
P =  atm
7. (b) Rate of diffusion ?
 Molecular mass of HCl > Molecular mass of NH
3
 HCl diffuses at slower rate and white ammonium chloride is
first formed near HCl bottle.
8. (b)    
Mean molar mass = 
= 
9. (a)
10. (c) According to Avogadro’s law "At same temperature and pressure, 
Volume ? no. of moles"
 = 
Q = =   = 16 : 1
: 2
11. (a) Given, 
According to Graham's law of diffusion for two different gases.
? x = 8
? Fraction of O
2
 = 1/8
12. (c) As temperature rises the most probable speed increases and the
fraction of molecules possessing most probable speed decreases.
13. (b) According to Boyle's law, PV = constant
?  log P + log V = constant
log P = – log V + constant
Hence, the plot of log P vs log V is straight line with negative slope.
14. (a)
Stoichoimetry ratio is 1 : 2
AT STP, P = 1 atm, T = 273 K, R = 0.0821
Page 4


1. (a) From the graph we can see the correct order of pressures  p
1
 > p
3 
>
p
2
2. (b) Gases become cooler during Joule Thomson’s expansion only if they are
below a certain temperature known as inversion temperature (T
i
). The
inversion temperature is characteristic of each gas and is given by
,   where R is gas constant
Given a = 0.244 atm L
2
 mol
–2
b = 0.027 L mol
–1
R = 0.0821 L atm deg
–1
 mol
–1
? 
3. (d)
4. (d) Let the mass of methane and oxygen = m gm.
Mole fraction of O
2
= 
=  =  = 
Partial pressure of O
2
 = Total pressure × mole fraction of O
2
, = P × 
 = 
5. (c) The expression of root mean square speed is
Hence,
6. (a) Given T = 27°C = 27 + 273 = 300 K
V = 10.0 L
Mass of He = 0.4 g
Mass of oxygen = 1.6 g
Mass of nitrogen = 1.4 g
n He = 0.4/4 = 0.1
n O
2
 = 1.6/32 = 0.05
n N
2
 = 1.4/28 = 0.05
n total = n He + n O
2
 + n N
2
 = 0.1 + 0.05 + 0.05 = 0.2
P =  atm
7. (b) Rate of diffusion ?
 Molecular mass of HCl > Molecular mass of NH
3
 HCl diffuses at slower rate and white ammonium chloride is
first formed near HCl bottle.
8. (b)    
Mean molar mass = 
= 
9. (a)
10. (c) According to Avogadro’s law "At same temperature and pressure, 
Volume ? no. of moles"
 = 
Q = =   = 16 : 1
: 2
11. (a) Given, 
According to Graham's law of diffusion for two different gases.
? x = 8
? Fraction of O
2
 = 1/8
12. (c) As temperature rises the most probable speed increases and the
fraction of molecules possessing most probable speed decreases.
13. (b) According to Boyle's law, PV = constant
?  log P + log V = constant
log P = – log V + constant
Hence, the plot of log P vs log V is straight line with negative slope.
14. (a)
Stoichoimetry ratio is 1 : 2
AT STP, P = 1 atm, T = 273 K, R = 0.0821
Initial moles of CO
2;
 n(CO
2
initial) 
= 0.022 mole
In final mixture no. of moles; n(CO
2
/CO mixture)
Increase in volume is by = 0.031 – 0.022
= 0.009 mole of gas
Final no. of moles of CO i.e. n
(CO final)
n
(CO final)
 = 2n
(CO2 initial)
 – n
(CO2 final)
= 2(0.022 – n
(CO2 final)
...(i)
n
(CO final)
 = 0.044 – 2n
(CO2 final)
...(ii)
? Now,  n
(CO final)
 + n
(CO final)
 = 0.031
n
(CO2 final)
 = 0.031 – n
(CO final)
...(ii)
Substituting (ii) in eq. (i)
n
(CO final)
 = 0.044 – 2[0.031 – n
(CO final)]
n
(CO final)
 = 0.044 – 0.062 + 2n
(CO final)
n
(CO final)
 = 0.018 mol. Volume of 
  = 0.40 Litre
and volume of CO
2
 = 0.7 litre – 0.4 litre
= 0.3 litre
?  CO
2 
= 300 mL, CO = 400 mL
15. (c) Most probable speed (C*) = 
Average Speed 
Page 5


1. (a) From the graph we can see the correct order of pressures  p
1
 > p
3 
>
p
2
2. (b) Gases become cooler during Joule Thomson’s expansion only if they are
below a certain temperature known as inversion temperature (T
i
). The
inversion temperature is characteristic of each gas and is given by
,   where R is gas constant
Given a = 0.244 atm L
2
 mol
–2
b = 0.027 L mol
–1
R = 0.0821 L atm deg
–1
 mol
–1
? 
3. (d)
4. (d) Let the mass of methane and oxygen = m gm.
Mole fraction of O
2
= 
=  =  = 
Partial pressure of O
2
 = Total pressure × mole fraction of O
2
, = P × 
 = 
5. (c) The expression of root mean square speed is
Hence,
6. (a) Given T = 27°C = 27 + 273 = 300 K
V = 10.0 L
Mass of He = 0.4 g
Mass of oxygen = 1.6 g
Mass of nitrogen = 1.4 g
n He = 0.4/4 = 0.1
n O
2
 = 1.6/32 = 0.05
n N
2
 = 1.4/28 = 0.05
n total = n He + n O
2
 + n N
2
 = 0.1 + 0.05 + 0.05 = 0.2
P =  atm
7. (b) Rate of diffusion ?
 Molecular mass of HCl > Molecular mass of NH
3
 HCl diffuses at slower rate and white ammonium chloride is
first formed near HCl bottle.
8. (b)    
Mean molar mass = 
= 
9. (a)
10. (c) According to Avogadro’s law "At same temperature and pressure, 
Volume ? no. of moles"
 = 
Q = =   = 16 : 1
: 2
11. (a) Given, 
According to Graham's law of diffusion for two different gases.
? x = 8
? Fraction of O
2
 = 1/8
12. (c) As temperature rises the most probable speed increases and the
fraction of molecules possessing most probable speed decreases.
13. (b) According to Boyle's law, PV = constant
?  log P + log V = constant
log P = – log V + constant
Hence, the plot of log P vs log V is straight line with negative slope.
14. (a)
Stoichoimetry ratio is 1 : 2
AT STP, P = 1 atm, T = 273 K, R = 0.0821
Initial moles of CO
2;
 n(CO
2
initial) 
= 0.022 mole
In final mixture no. of moles; n(CO
2
/CO mixture)
Increase in volume is by = 0.031 – 0.022
= 0.009 mole of gas
Final no. of moles of CO i.e. n
(CO final)
n
(CO final)
 = 2n
(CO2 initial)
 – n
(CO2 final)
= 2(0.022 – n
(CO2 final)
...(i)
n
(CO final)
 = 0.044 – 2n
(CO2 final)
...(ii)
? Now,  n
(CO final)
 + n
(CO final)
 = 0.031
n
(CO2 final)
 = 0.031 – n
(CO final)
...(ii)
Substituting (ii) in eq. (i)
n
(CO final)
 = 0.044 – 2[0.031 – n
(CO final)]
n
(CO final)
 = 0.044 – 0.062 + 2n
(CO final)
n
(CO final)
 = 0.018 mol. Volume of 
  = 0.40 Litre
and volume of CO
2
 = 0.7 litre – 0.4 litre
= 0.3 litre
?  CO
2 
= 300 mL, CO = 400 mL
15. (c) Most probable speed (C*) = 
Average Speed 
Root mean square velocity (C) =
16. (b) According to Graham’s Law Diffusion:
Since rate of diffusion =  ?    
or   
=  = 
 Mol. wt = 2 × V.D
? 
On calculating,
V
2
 = 14.1
17. (b) Compressibility factor 
(For one mole of real gas)
van der Waals equation 
At low pressure, volume is very large and hence correction term b can
be neglected in comparison to very large volume of V.
i.e.